In a class of $10$, there are $7$ students who play soccer. If the teacher chooses $3$ students, what is the probability that none of the three of them play soccer?
Solution: We can think about this problem as the probability of $3$ events happening. The first event is the teacher choosing one student who does not play soccer. The second event is the teacher choosing another student who does not play soccer, given that the teacher already chose someone who does not play soccer, and so on. The probabilty that the teacher will choose someone who does not play soccer is the number of students who do not play soccer divided by the total number of students: $\dfrac{3} {10}$ Once the teacher's chosen one student, there are only $9$ left. There's also one fewer student who does not play soccer, since the teacher isn't going to pick the same student twice. So, the probability that the teacher picks a second student who also does not play soccer is $\dfrac{2} {9}$ The probability of the teacher picking two students who do not play soccer must then be $\dfrac{3} {10} \cdot \dfrac{2} {9}$ We can continue using the same logic for the rest of the students the teacher picks. So, the probability of the teacher picking $3$ students such that none of them play soccer is $\dfrac{3}{10}\cdot\dfrac{2}{9}\cdot\dfrac{1}{8} = \dfrac{1}{120}$